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\begin{document}

\title{1-布朗运动的概念}
%\institute{上海立信会计金融学院}
\author{{\ppr LQW}}
\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
%\date{{\ppr 2022年12月29日} }

\maketitle

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\begin{frame}{内容提要 }

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\begin{enumerate}
\item[1.1.]  布朗运动的概念
\item[1.3.]  布朗运动的均值函数与协方差函数
\item[1.6.]  布朗桥、带漂移的布朗运动、几何布朗运动
\item[1.10.]  布朗运动的模拟、中心极限定理
\item[1.E.]  布朗运动的习题
\end{enumerate}

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\begin{itemize}%\itemsep1em

\item  {\color{red}定义：一个连续时间随机过程 $B=\{B_t,\, t\ge 0\}$ 称为是一个标准布朗运动，如果满足下述条件：}
\begin{enumerate}
\item  {\color{red}定义：这个随机过程从零开始，即 $B_0=0$. }
\item  {\color{red}定义：这个随机过程有平稳增量和独立增量。}
\item  {\color{red}定义：对任意 $t>0$, 随机变量 $B_t$ 服从正态分布 $N(0,t)$. }
\item  {\color{red}定义：这个随机过程的每条样本路径都是连续的。}
\end{enumerate}

\end{itemize}

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\begin{itemize}
\item  称随机过程 $\{X_t,\,\, t\ge 0\}$ 有平稳增量，是指对任意 $0\le s<t$ 与任意 $h>0$, 增量 $X_t - X_s$ 与增量 $X_{t+h} - X_{s+h}$ 有相同的分布。

\item  称随机过程 $\{X_t,\,\, t\ge 0\}$ 有独立增量，是指对任意 $0\le t_1 < t_2 < \cdots < t_n$, 下述增量是相互独立的，
$$X_{t_1},\, X_{t_2} - X_{t_1},\, \cdots, \, X_{t_n} - X_{t_{n-1}}. $$ 


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\begin{itemize}
\item  标准布朗运动 $B=\{B_t, t\ge 0\}$ 的均值函数为 $$\mu(t)=0.$$

\item  标准布朗运动 $B=\{B_t, t\ge 0\}$ 的协方差函数为 $$c(t,s)=cov(B_t,B_s) = \min(t,s). $$

%{\color{red}证明：}



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\begin{itemize}
\item  {\color{red}定义：一个随机过程 $\{X_t,\, t\ge 0\}$ 称为是 $h$-自相似的，如果对任意 $0\le t_1 < t_2 < \cdots < t_n$, 以及任意 $T>0$, 
下述两个随机向量有相同的分布，
$$( T^hX_{t_1},\, T^hX_{t_2},\, \cdots,\, T^hX_{t_n} ) \overset{d}{=}( X_{Tt_1},\, X_{Tt_2},\, \cdots,\, X_{Tt_n} ).$$
}

\item 标准布朗运动是 $0.5$-自相似的。

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\begin{itemize}
\item  布朗运动的样本路径是处处不可微的。

\item  布朗运动的样本路径在任意长度不为零的区间 $[0,T]$ 上的变差是无穷大，即 $$\sup\limits_{\tau} \sum\limits_{k=1}^{n} \lvert B_{t_i}(\omega) - B_{t_{i-1}}(\omega) \rvert =\infty. $$
这里 $\tau$ 取遍区间 $[0,T]$ 的所有分划：$0\le t_0 < t_1 < \cdots < t_n\le T$. 


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\begin{frame}{1.6. 布朗桥}

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\begin{itemize}
\item  {\color{red}定义：设 $B=\{B_t,\, t\ge 0\}$ 是标准布朗运动。下述随机过程 $X$ 称为布朗桥，$$X_t = B_t - tB_1,\, 0\le t\le 1. $$  }

\item  设 $X$ 为总体为区间 $[0,1]$ 上的均匀分布，设 $(X_1,X_2,\cdots,X_n)$ 是一个简单随机样本。
则经验分布函数 $F_n(x)$ 的极限是一个布朗桥。

\item  这是非参数统计的一个理论基础。

%\item  The Brownian bridge appears as the limit process of the normalized empirical distribution function of a sample of iid uniform $U(0,1)$ random variables. 

%\item  {\color{red}This is a fundamental result from non-parametric statistics; it is the basis for numerous goodness-of-fit tests in statistics. }

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\begin{itemize}
\item  {\color{red}定义：设 $B=\{B_t,\, t\ge 0\}$ 是标准布朗运动。带漂移的布朗运动是指随机过程
$$X_t = \mu t + \sigma B_t,\,\, t\ge 0.$$
}

\item  证明这是一个高斯过程。

\item  计算均值函数与协方差函数，$$\mu(t) = \mu t,\,\, c(t,s) = \sigma^2 \min(t,s). $$

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\begin{itemize}
\item   {\color{red}定义：设 $B=\{B_t,\, t\ge 0\}$ 是标准布朗运动。几何布朗运动是指随机过程 $$X_t = \exp\left( \mu t + \sigma B_t \right),\,\, t\ge 0. $$
}

\item  计算几何布朗运动的均值函数与协方差函数（设 $s\le t$）：
\begin{eqnarray*}
\mu (t) &=& \exp\left(\mu t + \frac{1}{2}\sigma^2 t \right), \\
c(t,s) &=& \exp\left[ \left( \mu + \frac{1}{2}\sigma^2 \right) (t+s) \right] \left[ \exp(\sigma^2s)-1 \right]. 
\end{eqnarray*}


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\begin{itemize}
%\item  The CLT says that the properly normalized and centered partial sums of an iid finite variance sequence {\color{red}converge in distribution} to a normal distribution. 

\item  中心极限定理是说，独立同分布的有限方差的随机变量的和，标准化之后依测度收敛到标准正态分布。


\item  设 $Y_1,Y_2,\cdots, Y_n,\cdots$ 是独立同分布的随机变量，均值为 $\mu$, 方差为 $\sigma^2$. 

\item  则如下定义的部分和过程的标准化依概率收敛于标准正态分布， 
$$R_0=0,\, R_n=Y_1+Y_2+ \cdots +Y_n, \, n\ge 1. $$

\item  如下定义的随机过程 $S_n$ 依概率收敛于标准布朗运动， 
\begin{eqnarray*}
S_n(t) = \left\{\begin{array}{ll}
\frac{R_i - i\mu}{\sigma\sqrt{n}}, & t=i/n, i=0,1,\cdots,n, \\
\text{linear interpolated}, & \text{elsewhere}.
\end{array}\right.
\end{eqnarray*}

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{\small\color{blue}
\begin{Verbatim}[numbers=left,xleftmargin=5mm]
import numpy as np
import matplotlib.pyplot as plt

N=50
Y=np.random.randn(N) #产生独立同分布的随机数
R=np.zeros(N+1)
R[1:N+1]=np.cumsum(Y) #得到部分和序列
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t=np.linspace(0,1,N+1)
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{\small\color{blue}
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fig=plt.figure()
ax=fig.add_subplot(111)

ax.hlines(y=0,xmin=-0.2,xmax=1.2)
ax.vlines(x=0,ymin=-2,ymax=2)

ax.plot(t,S,'b-')  #折线就是线性插值

ax.set_xlabel('t = time')
ax.set_ylabel('S(t)')
fig.savefig('bm-sample-path.png')
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\begin{itemize}

\item  %第1题：多项选择
A Stochastic process $B=\{B_t,\, t\in [0,\infty)\}$ is called standard Brownian motion or a Wiener process if the following conditions are satisfied.

\begin{enumerate}
\item[a.]  It starts at zero and it has continuous sample paths. 
\item[b.]  It has stationary increments, i.e., for all $t>s\ge 0$, and $h\ge 0$, the increment $X_t - X_s$ and $X_{t+h} - X_{s+h}$ have the same distribution. 
\item[c.]  It has independent increments, i.e., for all $n\ge 1$ and $t_n > \cdots > t_2 > t_1 \ge 0$, 
the increments $X_{t_2} - X_{t_1},  \cdots, X_{t_n} - X_{t_{n-1}}$ are independent random variables. 
\item[d.]  For every $t>0$, $B_t$ has a normal distribution $N(0,t)$. 
\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：abcd. 一个连续时间随机过程如果符合这些条件，那么称为标准布朗运动。

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\item  %第2题 
$B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. Find the distribution of the random variable $B_5-B_3$. 
\begin{enumerate}
\item[a.]  $N(0,2)$. 
\item[b.]  $N(0,3)$. 
\item[c.]  $N(0,5)$. 
\item[d.]  $N(0,8)$. 
\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：a. 增量 $B_5-B_3$ 与随机变量 $B_2$ 有同样的分布，即均值为零，方差为 2 的正态分布。

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\begin{itemize}
\item  %第3题 
Let $B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. Find the distribution of the random variable $B_5+B_3$. 
\begin{enumerate}
\item[a.]  $N(0,3)$. 
\item[b.]  $N(0,5)$. 
\item[c.]  $N(0,8)$. 
\item[d.]  $N(0,14)$. 
\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：d. 增量 $B_5$ 与 $B_3$ 不是独立的。因此首先将所求随机变量写成 $$B_5+B_3 = (B_5-B_3) + 2B_3,$$
其中根据布朗运动的独立增量性质，随机变量 $B_5-B_3$ 与 $B_3$ 是独立的。因为 $B_5-B_3\sim N(0,2)$, 而 $B_3\sim N(0,3)$ 故 $2B_3\sim N(0,12)$, 因此  $(B_5-B_3) + 2B_3$ 服从正态分布 $N(0,14)$. 

}


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\begin{itemize}
\item  %第4题 
Let $B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. Find the expectation function $\mu(t)=\mathbb{E}(B_t)$ and covariance function $c(t,s)=\mathbb{E}(B_tB_s)$. Here $t<s$. 
\begin{enumerate}
\item[a.]  $\mu(t)=0,\,\, c(t,s)=t$. 
\item[b.]  $\mu(t)=t,\,\, c(t,s)=t$. 
\item[c.]  $\mu(t)=0,\,\, c(t,s)=s$. 
\item[d.]  $\mu(t)=t,\,\, c(t,s)=s$. 
\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：b. 从 $B_t\sim N(0,t)$ 我们立刻知道均值函数 $\mu(t)=0$. 为计算协方差函数 $\mathbb{E}(B_tB_s)$, 因为 $t<s$, 所以我们首先将 $B_tB_s$ 写成 $B_t[B_t+(B_s-B_t)]$. 因为 $B_t$ 与 $B_s-B_t$ 相互独立，所以 $\mathbb{E}[B_t(B_s-B_t)] = 0$. 所以协方差函数为 $\mathbb{E}(B_t^2)=t.$

}


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\begin{itemize}
\item  %第5题 
Brownian motion is $0.5$-self-similar, i.e., for all $n\ge 1$ and $t_n > \cdots > t_2 > t_1 \ge 0$, for all $T>0$, 
the vector of random variables $\left( \sqrt{T}B_{t_1}, \sqrt{T}B_{t_2}, \cdots, \sqrt{T}B_{t_n} \right)$ has the same distribution as the vector of random variables $\left( B_{Tt_1}, B_{Tt_2}, \cdots, B_{Tt_n} \right)$. Find the correct statements. 
\begin{enumerate}
\item[a.]  The random variables $2B_2$ and $B_8$ have the same distribution. 
\item[b.]  The random vectors $(2B_1, 2B_2, 2B_3)$ and $(B_4, B_8, B_{12})$ have the same distribution. 
\item[c.]  The random variables $3B_1$ and $B_9$ are equal.  
\item[d.]  The sample path of a Brownian motion is nowhere differentiable. 
\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：abd. 随机变量 $3B_1$ 与 $B_9$ 有相同的分布，即 $N(0,9)$, 但是它们不相等。

}

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\begin{itemize}
\item  %第6题 
The variation of a function $f(t),0\le t\le T$ is defined as the supreme of the variations $$\sup_\tau \sum\limits_{k=1}^{n} \Big{|} f(t_k) - f(t_{k-1}) \Big{|}, $$
where the supremum is taken over all possible partitions $\tau: \,\, 0=t_0<t_1<\cdots<t_n=T$. 
Find the correct statements about variations of functions. 
\begin{enumerate}
\item[a.]  The variation of the function $f(t)=t^2$ over $[0,2]$ is $4$.
\item[b.]  The variation of the function $f(t)=t^2$ over $[-2,2]$ is $4$.
\item[c.]  The variation of the function $f(t)=\sin(t)$ over $[0,2\pi]$ is $4$.
\item[d.]  The variation of the Brownian motion $B_t$ over $[0,T]$ is infinity for any $T>0$. 
\end{enumerate} 

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\item  %第6题解答 
{\color{red}解答：acd. 对于单调递增或单调递减的函数 $f(t),t\in [a,b]$, 其变差为 $|f(b)-f(a)|$. 因此 $f(t)=t^2$ 在区间 $[0,2]$ 上的变差是 $4$, 在区间 $[-2,2]$ 上的变差是 $8$. 正弦函数 $\sin(x)$ 在区间 $[0,\frac{\pi}{2}]$ 上的变差是 $1$, 在区间 $[\frac{\pi}{2}, \pi]$ 上的变差也是 $1$, 如此可以计算在区间 $[0,2\pi]$ 上的变差为 $4$. 另一方面，布朗运动的路径在任何（长度大于零的）区间上的变差是无穷大。
}

\vspace{0.2cm}

\item  {\color{red} The unbounded variation and non-differentiability of Brownian sample paths are major reasons for the failure of classical integration methods, when applied to these paths, and for the introduction of stochastic calculus. 

}

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\begin{itemize}
\item  %第7题 
Let $B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. Consider the process $$X_t = B_t -tB_1, \,\, 0\le t\le 1. $$
Then $X_0=0, X_1=0$. For this reason this process has the name Brownian bridge. Find the correct statements. 
\begin{enumerate}
\item[a.]  The finite dimensional distributions of $X$ are multivariate Gaussian. 
\item[b.]  The expectation function of $X$ is $\mu(t)=0$. 
\item[c.]  The covariance function of $X$ is $c(t,s)=\min(t,s)-ts$, $s,t\in [0,1]$. 
\item[d.]  The Brownian bridge appears as the limit process of the normalized empirical distribution function of a sample of uniform random variables. 
\end{enumerate} 

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\begin{itemize}
\item  %第7题解答 
{\color{red}解答：abcd. 协方差函数需要计算。设 $t<s$, 于是 
\begin{eqnarray*}
c(t,s) &=& \mathbb{E}(X_tX_s) \\
&=& \mathbb{E}[(B_t -tB_1)(B_s -sB_1)] \\ 
&=& \mathbb{E}(B_tB_s -tB_1B_s - sB_1B_t + tsB_1^2) \\
&=& t - ts - st +ts \\
&=& t - ts. 
\end{eqnarray*}

}

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\begin{itemize}
\item  %第8题 
Let $B=\{B_t,\, t\in [0,\infty)\}$ be a standard Brownian motion. 
Consider the following two processes $Y_t=\mu t + \sigma B_t$ and $X_t = \exp(\mu t + \sigma B_t)$. 
Find the correct statements.
\begin{enumerate}
\item[a.]  The process $Y$ has the name Brownian motion with drift. 
\item[b.]  This process $X$ has the name Geometric Brownian motion. 
\item[c.]  Compute the expectation function of the process $X$, we see that 
$\mu(t) = \exp\left( \mu t+\sigma^2 t \right).$
\item[d.]  Compute the expectation function of the process $X$, we see that 
$\mu(t) = \exp\left( \mu t+\frac{1}{2}\sigma^2 t \right).$
\end{enumerate} 

\vspace{0.2cm}

\item  {\color{red}解答：abd. 对于标准正态分布 $Z$, 我们可以计算随机变量 $\exp(\lambda Z)$ 的数学期望
\begin{eqnarray*}
\mathbb{E}\left[ \exp(\lambda Z) \right] = \exp\left( \frac{1}{2}\lambda^2\right). 
\end{eqnarray*}

}

\end{itemize}

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